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Nov 18, 2021 at 16:12 comment added Michael Seifert @IlmariKaronen is correct. This approximation would only be valid if the distances from the balls to the center of the Earth ($r_A$, $r_B$, and $r_C$) were much less than 22 meters.
Nov 18, 2021 at 15:56 comment added Ilmari Karonen Isn't this backwards? The binomial approximation is valid for $x$ close to $0$, but in this case $\frac{r_A}{22\,{\rm m}} \gg 1$.
Nov 18, 2021 at 15:11 comment added Brendan Darrer @M-B Thanks, I corrected the typo!
S Nov 18, 2021 at 15:01 history edited Brendan Darrer CC BY-SA 4.0
improved answer
S Nov 18, 2021 at 15:01 history suggested Cross CC BY-SA 4.0
Improved formatting
Nov 18, 2021 at 14:04 review Suggested edits
S Nov 18, 2021 at 15:01
Nov 18, 2021 at 12:02 comment added AccidentalTaylorExpansion To make this approach more general if you have some condition like $r\ll R$ you can rewrite it as $\frac{r}{R}\equiv x\ll 1$. If you have a function $f(r,R)$ that you want to approximate you can replace each $r$ with $xR$ and finally you can Taylor expand in $x$ because $x$ is a small parameter (while taking $R$ constant)
Nov 18, 2021 at 7:51 comment added Mark H @M-B You are right about $22^2$.
Nov 18, 2021 at 5:13 comment added M-B Wait, I think there is a typo in your solution. Instead of $\frac{1}{\sqrt{22}}$ shouldn't it be $\frac{1}{22^2}$ ?
Nov 18, 2021 at 3:34 vote accept M-B
Nov 18, 2021 at 20:26
Nov 18, 2021 at 1:52 history edited Brendan Darrer CC BY-SA 4.0
improved grammar
Nov 18, 2021 at 1:26 history edited Brendan Darrer CC BY-SA 4.0
improved grammar
Nov 18, 2021 at 1:25 comment added M-B Thank you so much! I can't believe it didn't occur to me to take out the 22m. Kudos
Nov 18, 2021 at 1:23 vote accept M-B
Nov 18, 2021 at 3:34
Nov 18, 2021 at 1:22 history answered Brendan Darrer CC BY-SA 4.0